If the roots of the equation (z-4 )^3=8i are a-2i ,b+i and c+i, then abc is equal to

Question

Accepted Answer

Correct answer: C. (c) Given (z-4 )^3=8i ⇒z-4= (8i )^1/3 z=4+2 (i )^1/3 ⇒i=e^ iπ/2=e^( 2+2 nπ) ( ∵e^ iθ= cos ⁡ + isin ⁡ ) i^1/3=e^i( 6+ 2 nπ3) ,n=0 ,1 ,2 (e^i 6= cos 6+ isin 6= 32+i 12e^i( 6+ 2 3)=e^i 5 6= cos 5 6+ isin 5 6= - 32+i 12e^i( 6+ 4 3)=e^i 3 2= cos 3 2+ isin 3 2=0-i) Forn=0 ,e^i 6= 32+i 12 ⇒n=1 ,e^i 5 6=- 32+i 12 n=2 ,e^i 3 2=0-i ∴z=4+2( 32+i 12)= (4+ 3 )+i=b+i z=4+2( - 32+ i2)= (4- 3 )+i=c+i z=4+2 (0-i )=4-2i=a-2i ⇒a=4 ,b=4+ 3 andc=4- 3 ∴ abc= 4 (4+ 3 ) (4- 3 )= 4 (16-3 )=2 13

$If the roots of the equation (z - 4)^{3} = 8 i are a - 2 i, b + i and c + i, then abc is equal to$

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